4
$\begingroup$

This question already has an answer here:

Dimension analysis is a nice tool to create functions using physics dimensions that are desirable for our lives. I know, as an axiom, of sorts to me, that addition and subtraction must conserve units, and thus are actually relatively uncommon in physics equations.

Now, for why that is, I would tell someone it's the same reason why you only can add terms with the same qualities as that term in math as well (such as $x^2$ only being able to be added with scalar multiples of $x^2$, the same applying to $y^5$ or $ e^{-2x}\cosh(3x)$, for example). Except in the case of inputs, we have... (and here is where my explanation gets shady), vector units or magnitudes with magnitude 1 of things such as $\text{meters/second}$, $\text{Newtons}$, $\text{Tesla}$, $\text{Joules/Coulomb}$ etc.

However, I can't seem to necessarily explain why I couldn't say, add $x^2$ to $x$ or a velocity to a force other than my body simply not letting me out of habit, and my only explanation is: "You just can't", or "It'd look bizarre."

I need a better explanation for this. Could someone enlighten me? My two questions:

  • Why does adding and subtraction break down at things with different dimensions?

  • Why does multiplication/division allow for it?

$\endgroup$

marked as duplicate by Kyle Kanos, By Symmetry, Emilio Pisanty, John Rennie, Hritik Narayan Jun 2 '17 at 17:43

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Perhaps, and this was a bit useful, but I'm still a bit confused, especially about the isomorphism bit - also, the second question wasn't answered too directly in that. $\endgroup$ – sangstar Jun 2 '17 at 17:03
  • $\begingroup$ Multiplication/division can be explained because most of the relation between different quantities are derived from the proportionality or observation, eg. in capacitor $\text{q}\propto\text{V}\implies\text{q}=\text{CV}$. We just keep using them by substituting these quantities to another which consequently make it happen. $\endgroup$ – Saharsh Jun 2 '17 at 17:27
  • $\begingroup$ In the future, maybe we find a new force $F$ which satisfies the following law $F = \alpha (r + m_1 m_2)$. So, in the future maybe we can add different quantities. $\endgroup$ – tchappy ha Feb 1 at 2:38
0
$\begingroup$

I'm not sure I really understand what you're getting at but here goes: Say you add 2 metres to 3 kg - what have you got? I'd say you have just got 2 metres and 3 kg, which is not telling you anything. You can't say you have 5 of something, although if the metres and kilograms were apples and oranges, respectively, you could say you had 5 pieces of fruit.

$\endgroup$
  • $\begingroup$ But, essentially, why is that adding 2 meters to 3 kg gives you "2 meters and 3 kg", but when multiplying you have $6 meters\ kg$? Why can't you have $5 meters\ kg$? Why is this the mechanism? $\endgroup$ – sangstar Jun 2 '17 at 17:49
  • 1
    $\begingroup$ Adding 2m and 3kg gives you something we can call a "(2,3) m,kg". We can't represent that as a "5 m,kg" because we then cannot differentiate it from a "(1,4) m,kg" which is 1m added to 4kg. Clearly 1m added to 4kg is not the same as 2m added to 3kg. What if you told me to subtract a "2 m,kg" from a "5 m,kg" ? Am I to take the 2 from the meters part or the kg part ? Do I distribute it between both parts ? There's no way to be consistent doing this. $\endgroup$ – StephenG Jun 2 '17 at 18:21

Not the answer you're looking for? Browse other questions tagged or ask your own question.